Bunuel wrote:
We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).
The probability that the point will be from this region is: Area of this region/Area of the triangle.
Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.
P=8/10=4/5
Answer: E.
If i go by the way u say....
shouldn't the probability be <4/5?
probability = area of region where x-y>0 / total are of region = x/0.5*4*5 = x/10
point (4,4) is where x-y=0. point(4, <4) is where x-y>0.
x = 0.5 * 4*(<4) = <8 hence x/10 = <8/10 = <4/5
the point here is that to get 4/5 you would have to take the region where x and y = 4 in which case x-y is not >0. any value of y below 4 would give x-y>0 hence the probability should be less than 4/5...I dont think you can place an exact value on the probability as x and y can be decimals...